Let `A={x inR: x >=1/2} and B={x in R: x>=3/4}.` If `f:A->B` is defined as `f(x)=x^2-x=1,` then the solution set of the equation `f(x)=f^-1(x)` is

Options

A. {1}
B. {2}
C. {1//2}
D. None of these

Explanation

To solve the problem, we need to find the solution set of the equation \( f(x) = f^{-1}(x) \) given the function \( f(x) = x^2 - x + 1 \) for \( x \in A \) and \( f^{-1}(x) \).

### Step-by-Step Solution:

1. **Define the function and its domain:**
We have \( A = \{ x \in \mathbb{R} : x \geq \frac{1}{2} \} \) and \( B = \{ x \in \mathbb{R} : x \geq \frac{3}{4} \} \). The function is defined as:
\[ f(x) = x^2 - x + 1 \]

2. **Find the inverse function \( f^{-1}(x) \):**
To find \( f^{-1}(x) \), we start by setting \( y = f(x) \):
\[ y = x^2 - x + 1 \]
Rearranging gives:
\[ x^2 - x + (1 - y) = 0 \]
This is a quadratic equation in \( x \). We can use the quadratic formula:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (1 - y)}}{2 \cdot 1} \]
Simplifying this gives:
\[ x = \frac{1 \pm \sqrt{4y - 3}}{2} \]
Since \( f(x) \) is increasing in the domain \( A \), we take the positive root:
\[ f^{-1}(x) = \frac{1 + \sqrt{4x - 3}}{2} \]

3. **Set up the equation \( f(x) = f^{-1}(x) \):**
We need to solve:
\[ x^2 - x + 1 = \frac{1 + \sqrt{4x - 3}}{2} \]

4. **Clear the fraction:**
Multiply both sides by 2:
\[ 2(x^2 - x + 1) = 1 + \sqrt{4x - 3} \]
Simplifying gives:
\[ 2x^2 - 2x + 2 - 1 = \sqrt{4x - 3} \]
Thus:
\[ 2x^2 - 2x + 1 = \sqrt{4x - 3} \]

5. **Square both sides to eliminate the square root:**
\[ (2x^2 - 2x + 1)^2 = 4x - 3 \]

6. **Expand the left side:**
\[ 4x^4 - 8x^3 + 4x^2 + 4x - 4x^2 + 1 = 4x - 3 \]
Simplifying gives:
\[ 4x^4 - 8x^3 + 1 + 3 = 0 \]
Thus:
\[ 4x^4 - 8x^3 + 4 = 0 \]

7. **Factor out the common term:**
\[ 4(x^4 - 2x^3 + 1) = 0 \]
This simplifies to:
\[ x^4 - 2x^3 + 1 = 0 \]

8. **Solve the polynomial equation:**
Using the Rational Root Theorem or synthetic division, we can find that \( x = 1 \) is a root. Factoring gives:
\[ (x - 1)^2(x^2 - 1) = 0 \]
Thus:
\[ (x - 1)^2(x - 1)(x + 1) = 0 \]
The roots are \( x = 1 \) (with multiplicity 2) and \( x = -1 \).

9. **Check the validity of the solutions:**
Since \( x \) must be in the domain \( A \) where \( x \geq \frac{1}{2} \), we discard \( x = -1 \). The only valid solution is:
\[ x = 1 \]

### Final Solution:
The solution set of the equation \( f(x) = f^{-1}(x) \) is:
\[ \{ 1 \} \]